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**Sample Exercise 6.1 Concepts of Wavelength and Frequency**

Two electromagnetic waves are represented in the margin. (a) Which wave has the higher frequency? (b) If one wave represents visible light and the other represents infrared radiation, which wave is which? Solution (a) Wave 1 has a longer wavelength (greater distance between peaks). The longer the wavelength, the lower the frequency (ν = c ⁄ λ). Thus, Wave 1 has the lower frequency, and Wave 2 has the higher frequency. (b) The electromagnetic spectrum (Figure 6.4) indicates that infrared radiation has a longer wavelength than visible light. Thus, Wave 1 would be the infrared radiation.

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**Sample Exercise 6.1 Concepts of Wavelength and Frequency**

Continued Practice Exercise 1 A source of electromagnetic radiation produces infrared light. Which of the following could be the wavelength of the light? (a) 3.0 nm (b) 4.7 cm (c) 66.8 m (d) 34.5 μm (e) 16.5 Å Practice Exercise 2 If one of the waves in the margin represents blue light and the other red light, which wave is which?

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**Sample Exercise 6.2 Calculating Frequency from Wavelength**

The yellow light given off by a sodium vapor lamp used for public lighting has a wavelength of 589 nm. What is the frequency of this radiation? Solution Analyze We are given the wavelength, λ, of the radiation and asked to calculate its frequency, ν. Plan The relationship between the wavelength and the frequency is given by Equation 6.1. We can solve for ν and use the values of λ and c to obtain a numerical answer. (The speed of light, c, is 3.00 × 108 m ⁄ s to three significant figures.) Solve Solving Equation 6.1 for frequency gives (ν = c ⁄ λ). When we insert the values for c and λ, we note that the units of length in these two quantities are different. We can convert the Wavelength from nanometers to meters, so the units cancel: Check The high frequency is reasonable because of the short wavelength. The units are proper because frequency has units of “per second,” or s–1.

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**Sample Exercise 6.2 Calculating Frequency from Wavelength**

Continued Practice Exercise 1 Consider the following three statements: (i) For any electromagnetic radiation, the product of the wavelength and the frequency is a constant. (ii) If a source of light has a wavelength of 3.0 Å, its frequency is 1.0 × 1018 Hz. (iii) The speed of ultraviolet light is greater than the speed of microwave radiation. Which of these three statements is or are true? (a) Only one statement is true. (b) Statements (i) and (ii) are true. (c) Statements (i) and (iii) are true. (d) Statements (ii) and (iii) are true. (e) All three statements are true. Practice Exercise 2 (a) A laser used in orthopedic spine surgery produces radiation with a wavelength of 2.10 μm. Calculate the frequency of this radiation. (b) An FM radio station broadcasts electromagnetic radiation at a frequency of MHz (megahertz; 1 MHz = 106 s–1). Calculate the wavelength of this radiation. The speed of light is × 108 m ⁄ s to four significant figures.

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**Sample Exercise 6.3 Energy of a Photon**

Calculate the energy of one photon of yellow light that has a wavelength of 589 nm. Solution Analyze Our task is to calculate the energy, E, of a photon, given λ = 589 nm. Plan We can use Equation 6.1 to convert the wavelength to frequency: ν = c ⁄ λ We can then use Equation 6.3 to calculate energy: E = hν Solve The frequency, ν, is calculated from the given wavelength, as shown in Sample Exercise 6.2: ν = c ⁄ λ = 5.09 × 1014 s–1 The value of Planck constant, h, is given both in the text and in the table of physical constants on the inside back cover of the text, and so we can easily calculate E: E = (6.626 × 10–34 J-s)(5.09 × 1014 s–1) = 3.37 × 10–19 J Comment If one photon of radiant energy supplies 3.37 × 10–19 J, we calculate that one mole of these photons will supply: (6.02 × 1023 photons ⁄ mol)(3.37 × 10–19 J ⁄ photon) = 2.03 × 105 J ⁄ mol

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**Sample Exercise 6.3 Energy of a Photon**

Continued Practice Exercise 1 Which of the following expressions correctly gives the energy of a mole of photons with wavelength λ? Practice Exercise 2 (a) A laser emits light that has a frequency of 4.69 × 1014 s–1. What is the energy of one photon of this radiation? (b) If the laser emits a pulse containing 5.0 × 1017 photons of this radiation, what is the total energy of that pulse? (c) If the laser emits 1.3 × 10–2 J of energy during a pulse, how many photons are emitted?

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**Sample Exercise 6.7 Orbital Diagrams and Electron Configurations**

Draw the orbital diagram for the electron configuration of oxygen, atomic number 8. How many unpaired electrons does an oxygen atom possess? Solution Analyze and Plan Because oxygen has an atomic number of 8, each oxygen atom has eight electrons. Figure 6.25 shows the ordering of orbitals. The electrons (represented as half arrows) are placed in the orbitals (represented as boxes) beginning with the lowest-energy orbital, the 1s. Each orbital can hold a maximum of two electrons (the Pauli exclusion principle). Because the 2p orbitals are degenerate, we place one electron in each of these orbitals (spin-up) before pairing any electrons (Hund’s rule). Solve Two electrons each go into the 1s and 2s orbitals with their spins paired. This leaves four electrons for the three degenerate 2p orbitals. Following Hund’s rule, we put one electron into each 2p orbital until all three orbitals have one electron each. The fourth electron is then paired up with one of the three electrons already in a 2p orbital, so that the orbital diagram is The corresponding electron configuration is written 1s22s22p4. The atom has two unpaired electrons.

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**Sample Exercise 6.7 Orbital Diagrams and Electron Configurations**

Continued Practice Exercise 1 How many of the elements in the second row of the periodic table (Li through Ne) will have at least one unpaired electron in their electron configurations? (a) 3 (b) 4 (c) 5 (d) 6 (e) 7 Practice Exercise 2 (a) Write the electron configuration for silicon, element 14, in its ground state. (b) How many unpaired electrons does a ground-state silicon atom possess?

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**Sample Exercise 6.8 Electron Configurations for a Group**

What is the characteristic valence electron configuration of the group 7A elements, the halogens? Solution Analyze and Plan We first locate the halogens in the periodic table, write the electron configurations for the first two elements, and then determine the general similarity between the configurations. Solve The first member of the halogen group is fluorine (F, element 9). Moving backward from F, we find that the noble-gas core is [He]. Moving from He to the element of next higher atomic number brings us to Li, element 3. Because Li is in the second period of the s block, we add electrons to the 2s subshell. Moving across this block gives 2s2. Continuing to move to the right, we enter the p block. Counting the squares to F gives 2p5. Thus, the condensed electron configuration for fluorine is F: [He]2s22p5 The electron configuration for chlorine, the second halogen, is Cl: [Ne]3s23p5 From these two examples, we see that the characteristic valence electron configuration of a halogen is ns2np5, where n ranges from 2 in the case of fluorine to 6 in the case of astatine.

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**Sample Exercise 6.8 Electron Configurations for a Group**

Continued Practice Exercise 1 A certain atom has an ns2np6 electron configuration in its outermost occupied shell. Which of the following elements could it be? (a) Be (b) Si (c) I (d) Kr (e) Rb Practice Exercise 2 Which family of elements is characterized by an ns2np2 electron configuration in the outermost occupied shell?

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**Sample Exercise 6.9 Electron Configurations from the Periodic Table**

(a) Based on its position in the periodic table, write the condensed electron configuration for bismuth, element 83. (b) How many unpaired electrons does a bismuth atom have? Solution (a) Our first step is to write the noble-gas core. We do this by locating bismuth, element 83, in the periodic table. We then move backward to the nearest noble gas, which is Xe, element 54. Thus, the noble-gas core is [Xe]. Next, we trace the path in order of increasing atomic numbers from Xe to Bi. Moving from Xe to Cs, element 55, we find ourselves in period 6 of the s block. Knowing the block and the period identifies the subshell in which we begin placing outer electrons, 6s. As we move through the s block, we add two electrons: 6s2. As we move beyond the s block, from element 56 to element 57, the curved arrow below the periodic table reminds us that we are entering the f block. The first row of the f block corresponds to the 4f subshell. As we move across this block, we add 14 electrons: 4f 14. With element 71, we move into the third row of the d block. Because the first row of the d block is 3d, the second row is 4d and the third row is 5d. Thus, as we move through the ten elements of the d block, from element 71 to element 80, we fill the 5d subshell with ten electrons: 5d10.

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**Sample Exercise 6.9 Electron Configurations from the Periodic Table**

Continued Moving from element 80 to element 81 puts us into the p block in the 6p subshell. (Remember that the principal quantum number in the p block is the same as that in the s block.) Moving across to Bi requires three electrons: 6p3. The path we have taken is

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**Sample Exercise 6.9 Electron Configurations from the Periodic Table**

Continued Putting the parts together, we obtain the condensed electron configuration: [Xe]6s24f 145d106p3. This configuration can also be written with the subshells arranged in order of increasing principal quantum number: [Xe]4f 145d106s26p3. Finally, we check our result to see if the number of electrons equals the atomic number of Bi, 83: Because Xe has 54 electrons (its atomic number), we have = 83. (If we had 14 electrons too few, we would realize that we have missed the f block.) (b) We see from the condensed electron configuration that the only partially occupied subshell is 6p. The orbital diagram representation for this subshell is In accordance with Hund’s rule, the three 6p electrons occupy the three 6p orbitals singly, with their spins parallel. Thus, there are three unpaired electrons in the bismuth atom. Practice Exercise 1 A certain atom has an [noble gas]5s24d105p4 electron configuration. Which element is it? (a) Cd (b) Te (c) Sm (d) Hg (e) More information is needed Practice Exercise 2 Use the periodic table to write the condensed electron configuration for (a) Co (element 27), (b) In (element 49).

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**Sample Integrative Exercise Putting Concepts Together**

Boron, atomic number 5, occurs naturally as two isotopes, 10B and 11B, with natural abundances of 19.9% and 80.1%, respectively. (a) In what ways do the two isotopes differ from each other? Does the electronic configuration of 10B differ from that of 11B? (b) Draw the orbital diagram for an atom of 11B. Which electrons are the valence electrons? (c) Indicate three major ways in which the 1s electrons in boron differ from its 2s electrons. (d) Elemental boron reacts with fluorine to form BF3, a gas. Write a balanced chemical equation for the reaction of solid boron with fluorine gas. (e) ΔHf° for BF3(g) is − kJ ⁄ mol. Calculate the standard enthalpy change in the reaction of boron with fluorine. (f) Will the mass percentage of F be the same in 10BF3 and 11BF3? If not, why is that the case? Solution (a) The two isotopes of boron differ in the number of neutrons in the nucleus (Sections 2.3 and 2.4) Each of the isotopes contains five protons, but 10B contains five neutrons, whereas 11B contains six neutrons. The two isotopes of boron have identical electron configurations, 1s22s22p1, because each has five electrons. (b) The complete orbital diagram is 1s 2s 2p The valence electrons are the ones in the outermost occupied shell, the 2s2 and 2p1 electrons. The 1s2 electrons constitute the core electrons, which we represent as [He] when we write the condensed electron configuration, [He]2s22p1.

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**Sample Integrative Exercise Putting Concepts Together**

Continued (c) The 1s and 2s orbitals are both spherical, but they differ in three important respects: First, the 1s orbital is lower in energy than the 2s orbital. Second, the average distance of the 2s electrons from the nucleus is greater than that of the 1s electrons, so the 1s orbital is smaller than the 2s. Third, the 2s orbital has one node, whereas the 1s orbital has no nodes (Figure 6.19).

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**Sample Integrative Exercise Putting Concepts Together**

Continued (d) The balanced chemical equation is 2 B(s) + 3 F2(g) → 2 BF3(g) (e) ΔH ° = 2(−1135.6) − [0 + 0] = − kJ. The reaction is strongly exothermic. (f) As we saw in Equation 3.10 (Section 3.3), the mass percentage of an element in a substance depends on the formula weight of the substance. The formula weights of 10BF3 and 11BF3 are different because of the difference in the masses of the two isotopes (the isotope masses of 10B and 11B are and amu, respectively). The denominators in Equation 3.10 would therefore be different for the two isotopes, whereas the numerators would remain the same.

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Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J.

Copyright ©2009 by Pearson Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Introductory Chemistry, Third Edition By Nivaldo J.

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